Looking for the solution for this problem. The right answer is provided below. T

Question

Looking for the solution for this problem. The right answer is provided below. Thank you! A mixture of 0.75 mol of N2 and 1.20 mol H2 are placed in a 3.0 L container with an iron catalyst and heated under pressure to 750 K. N2 (g) 3H2(g) 2NH, (g) a. When the reaction reaches equilibrium, [H2] 0.10 M. What is the value of [N2] and [NH3] at equilibrium? b. If 0.30 mol of ammonia are removed through condensation and no other gases are added, what are the new [N2], [H2], and [NH, when the reaction re-establishes equilibrium?

Explanation / Answer

a) Molarity = moles/ volume in Litres

Given that

[N2] = 0.75 mol/ 3 L = 0.25 M

[H2] = 1.2 mol/ 3 M = 0.4 M

                N2       +    3H2    <----------------> 2NH3

Initial                   0.25 M      0.4 M                            0

At equilibrium        0.25 -x       0.4 -3x                         2x

Hence,

[H2]eq = 0.4-3x

But , given that [H2]eq = 0.1 M

Then,

     0.4-3x = 0.1

       x = 0.1 M

Therefore,

Equilbrium concentrations are

[N2] = 0.25- x = 0.25 M - 0.1 M = 0.15 M

[H2] = 0.1 M

[NH3] = 2x = 2 x 0.1 M = 0.2 M

  

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